Question

factor 4.9t^2-10t+2.5

Answer 1

using quadratic formula?

Answer 2

yes

Answer 3

Certainly you can factor using the quadratic formula, but I don't see that formula as being necessary. Note that 0.49 is the square of 0.7.

Answer 4

\[10\pm \sqrt{10}/18\]

Answer 5

Well,
Given the general form
\[a{x}^{2}+bx+c=0 \]

there exists two solutions where:
\[x=\frac{-b+\sqrt{{b}^{2}-4ac}}{2a},\frac{-b-\sqrt{{b}^{2}-4ac}}{2a}\]

In this case, a=4.9, b=−10 and c=2.5

\[x=\frac{-(-10) \pm\sqrt{{(-10)}^{2}-4(4.9)(2.5)}}{2(4.9)}\]

Answer 6

it would be \[t=\frac{10+\sqrt{{10}^{2}-49}}{9.8},\frac{10-\sqrt{{10}^{2}-49}}{9.8}\]

Answer 7

ok. so t= \[10\pm \sqrt{51}\div9.8\]

Answer 8

Close. You must (absolutely must) enclose that 10 plus or minus Sqrt(51) inside parentheses.

How would you check your answers?

Answer 9

would i plug that into the equation for x?

Answer 10

yes, because there are two solutions.